If we knew exactly how the price was moving... - page 4
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is wrong in principle here!
0<p<1 is probability
tp, sl are "kilos"
you can't put them in the same key
why not? You don't like points and tp,sl - play a game: bet a dollar each. You guess get yours and 2 more on top, you lose only the bet. Probability to guess or not to guess 0.5/0.5.
Mo=0.5*2-0.5*1=0.5. I.e. on average in each game you win 0.5 quid.
But it would be better to count pips if you don't consider MM yet, which is almost equivalent to MM with fixed lot.
How else do you calculate the mathematical expectation?
The mathematical expectation of a discrete distribution
- If X is a discrete random variable having a distribution
,then it follows directly from the definition of the Lebesgue integral that
. https://ru.wikipedia.org/wiki/%D0%9C%D0%B0%D1%82%D0%B5%D0%BC%D0%B0%D1%82%D0%B8%D1%87%D0%B5%D1%81%D0%BA%D0%BE%D0%B5_%D0%BE%D0%B6%D0%B8%D0%B4%D0%B0%D0%BD%D0%B8%D0%B5For example, we have an asymmetric distribution with mo=0. If it is asymmetric, it is possible to find a value of sl and tp, at which the new distribution will be with mo different from zero.
Similarly, for some symmetric but non-Gaussian distributions. By purely varying sl and tp
This statement does not correspond to reality.
It is known that for any first-difference series (FDS) distribution with MO=0, the introduction of sl and tp does not shift the expectation in any way. This is also true for asymmetric distributions.
Suppose we have a price series obtained by integration of a normally distributed SV with MO=0. Let us follow the strategy "let profits grow and cut losses". It is clear that we deal with "pure" martingale on which, as we know, no profitable strategy can be built (as well as loss-making one). Losses will be cut by a fixed Stop Loss and Take will be flexible and let's see how this parameter will change the MO of our TS.
The picture to the left shows distribution of losses for such TS with infinite Take (it simply does not exist). We can see that the distribution is essentially asymmetrical with long tails of positive takes (we let profits grow) and cut losses (the loss boundary is not sharp because of slippages). There are 4500 transactions in the experiment. The MO differs from zero by 7% of the typical size of the bribe, i.e. almost zero, which was expected (if more transactions are taken, zero will be more accurate).
Introduce the take. In the figure on the right it is about 10 times the average size of the payoff - MO has not moved (still 7%). On the right we can see a small tail that has grown around the mark, it is clear - we cut long distribution tails with the mark. Further, we bring the TP closer:
In the figure below on the left, the TP is equal to five middle plucks and two in the figure on the right. The outgrowth of the tail on the tp side is clearly visible.
It can be seen that the MO for the non-symmetric distribution has not changed.
All the above is also true for an integrated CB with a non-Gaussian distribution in the RPR, in particular for price series. The introduction of StopLoss and TakeProfit in TS does not change the TS yield (does not shift the MO), but only insures it from force majeure situations such as connection failure, etc.
P.S. For the classical definition of IR: If the probability density function F is known for some value x, then its average value is calculated as follows:
How else do you calculate the mathematical expectation?
The mathematical expectation of a discrete distribution
- If X is a discrete random variable having a distribution
,then it follows directly from the definition of Lebesgue integral that
.Go beyond wikipedia and look into What is an event, What is the probability of an event occurring, What is probability addition, etc.
Feller or Verlang or Shirochin or Wentzel ...
This statement does not correspond to reality.
It is known that for any first-difference series (FDS) distribution with MO=0, the introduction of sl and tp in no way shifts the expectation. This is also true for asymmetric distributions.
Suppose we have a price series obtained by integration of a normally distributed SV with MO=0. Let us follow the strategy "let profits grow and cut losses". It is clear that we deal with "pure" martingale on which, as we know, no profitable strategy can be built (as well as loss-making one). We will cut losses by a fixed Stop Loss and make Take a mobile one, and see how the MO of our TS will change.
The figure on the left shows distribution of losses for such TS with infinite Take (it simply does not exist). We can see that the distribution is essentially asymmetrical with long tails of positive takes (we let profits grow) and cut losses (the loss boundary is not sharp because of slippage). There are 4500 transactions in the experiment. The MO differs from zero by 7% of the typical size of the bribe, i.e. almost zero, which was expected (if more transactions are taken, zero will be more accurate).
Let's introduce the take. In the figure on the right it is about 10 times the average size of the payoff - MO has not moved (still 7%). On the right we can see a small tail that has grown around the tee, it is understandable - we cut long distribution tails with the tee. Let's bring the TP closer:
In the picture below, TP=5 middle take on the left and two on the right. You can clearly see the outgrowth of the tail on the tp side.
You can see that the MO for the non-symmetric distribution has not changed.
so you originally took a distribution of HP increments with mo=0. In this case, no introduction of stops and tokes will lead to positive mo.
This is all true for integrated CB with non-Gaussian distribution in RRR, in particular for price series. The introduction of StopLoss and TakeProfit into the TS does not change the TS yield (does not shift the MO), but only insures it against force majeure situations such as breakdowns, etc.
Of course, because the increments of the real series are symmetrical.
My Take Profit does not insure anything, but changes the MO and significantly :)
But it is asymmetric. That's what I was getting at. I highlighted the point in your post above.
you have asymmetric distributions obtained by varying sl and tp on an integrated distribution of normally distributed increments. This is how it should be. You are still trading a symmetric distribution and no way of varying sl and tp can make a positive mo.
Maybe I didn't express it precisely, but I meant the asymmetric distribution, by integrating which the series in question is obtained.
Of course, because the increments of the real series are symmetrical.
With me, take profit does not insure anything, but changes the MO and significantly :)
I'm talking about the distribution of bribe increments - it's asymmetric in my example and the introduction of a TR doesn't change anything, and that doesn't agree with your statement above.